STA-301 Assignment No.1 spring 2012 idea solution on 6th April 2012

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Question No.1



























48 29 31 32 54 33 44 36 38 31 46 30 20 44 47 39 42 35 33 47 31 35 34 42 41 42 43 35 32 35 43 36 37 45 46 41 25 27 26 40 38 41 44 47 45 45 52 43 44 43









































Classes Frequency C.Boundaries X



20 24 1 19.5 24.5 34.25



25 29 4 24.5 29.5 41.75



30 34 9 29.5 34.5 49.25



35 39 10 34.5 39.5 56.75



40 44 15 39.5 44.5 64.25



45 49 9 44.5 49.5 71.75



50 54 2 49.5 54.5 79.25






















X Mid points







F Frequency


























Question No.2






































Class Limits F C.Boundaries X FX


0.7312 0.7313 10 0.73115 0.73135 0.73125 7.3125


0.7314 0.7314 15 0.73135 0.73155 0.73145 10.97175


0.7316 0.7317 20 0.73155 0.73175 0.73165 14.633


0.7318 0.7319 25 0.73175 0.73195 0.73185 18.29625


0.732 0.7321 30 0.73195 0.73215 0.73205 21.9615


0.7322 0.7323 8 0.73215 0.73235 0.73225 5.858


0.7324 0.7325 2 0.73235 0.73255 0.73245 1.4649














110

5.12295 80.4979












EFX 80.4979







EF 110



























Mean = EFX








EF

















Mean= 80.4979








110

















Mean= 0.73179909

















Your means lies in the class boundary of 0.73175-0.73195














Mode


































Mode= L + Fm-F1 x h







(Fm-F1) + (Fm- F2)


































L Lower Class Boundary of the modal Class





Fm Maximum Frequency






F1 frequency of the class preceding the Modal Class




F2 frequency of the class Following the Modal Class




h Length of class difference





































Mode= 0.73195+ 30-25 x 0.0002







(30-25)+ (30-2)




































Mode= 0.731973


























You can clealry see that your mode 0.731973 lies in the modal class